Conference on Positive Polynomials,

Luminy March 14-18, 2005.

August 17, 2005

Prof. Dr. habil. Salma Kuhlmann ^{1} , Research Unit Algebra and Logic, University of Saskatchewan, McLean Hall, 106 Wiggins Road, Saskatoon, SK S7N 5E6, Canada

email: skuhlman@math.usask.ca

homepage: http://math.usask.ca/˜ skuhlman/index.html

1Partially supported by the Natural Sciences and Engineering Research Council of Canada.

The Invariant Moment Problem.

As all roads lead to Rome so I find in my own case at least that all algebraic inquiries, sooner or later, end at the Capitol of modern algebra over whose shining portal is inscribed the Theory of Invariants.

– J. J. Sylvester 1854

The slides of this talk are available at: http://math.usask.ca/˜ skuhlman/slideimp.pdf

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Abstract.

Let group G together with ϕ : G → GL_{n}(ℝ) a linear representation. Assume that G is such that the ring ℝ[X_{1}, ⋯, X_{n}]^{G} of G-invariant polynomials is a finitely generated subalgebra of the polynomial ring ℝ[X_{1}, ⋯, X_{n}]. We analyze the preorderings of ℝ[X_{1}, ⋯, X_{n}]^{G} associated to a Ginvariant semi-algebraic set. We formulate a G-invariant version of Haviland’s theorem concerning the representation of linear functinals by integrals. We exploit the correspondence between G-invariant semi-algebraic sets, and semi-algebraic sets in the orbit space, to solve the G-invariant moment problem. We produce many examples of closed unbounded G-invariant subsets K of ℝ^{n} for which the K-moment problem is not finitely solvable, but the G-invariant moment problem is finitely solvable. For these examples, G- invariant linear functionals are representable by an invariant measure supported on K, even though such a representation fails for arbitrary functionals.

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Plan of the talk.

1. Positive Polynomials and Invariant Theory.

2. Preorderings of the ring of invariant polynomials.

3. Semi-Algebraic Geometry in the Orbit Space.

4. Saturation.

5. G-Invariant Moment Problem.

6. The averaged Moment Problem.

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Positive Polynomials and Invariant Theory.

Let ℝ[X] := ℝ[X_{1}, ⋯, X_{n}], the polynomial ring in n variables. T ⊆ ℝ[X] is a preordering if f^{2} ∈ T, ∀f ∈ ℝ[X] and T is closed under addition and multiplication.

Given a subset S of A, there is a smallest preordering T_{S} containing S; the preordering generated by S:

T_{S} = {_{e∈{0,1}^{r}}^{∑} σ_{e}f^{e} : r ≥ 0, σ_{e} ∈ ∑ ℝ[X]^{2}, f_{1}, ⋯, f_{r} ∈ S}

where f^{e} := f_{1}^{e_{1}} ⋯ f_{r}^{e_{r}}, if e = (e_{1}, ⋯, e_{r}), and ∑ ℝ[X]^{2} denotes the sums of squares of ℝ[X].

Let S = {f_{1}, ⋯, f_{n}} ⊂ ℝ[X], S defines a basic closed semialgebraic subset of ℝ^{n}:

K = K_{S} = {x ∈ ℝ^{n} : f_{1}(x) ≥ 0, …, f_{s}(x) ≥ 0}.

We define the saturation of T_{S}:

Psd(K_{S}) := {f ∈ ℝ[X] : f ≥ 0 on K_{S}},

Psd(K_{S}) is a preordering in ℝ[X]. T_{S} is saturated if Psd(K_{S}) = T_{S}.

While considering the moment problem, we are interested in linear functinals L defined on the algebra ℝ[X], and non-negative on T_{S}. In particular, we work with the following corresponding preordering of ℝ[X]:

cl(T_{S}) := {f; L(f) ≥ 0 for all L ≠ 0 such that L(T_{S}) ≥ 0}.

cl(T_{S}) is the closure of T_{S} in ℝ[X]. We say that T_{S} is closed if cl(T_{S}) = T_{S}. We have the inclusions

T_{S} ⊆ cl(T_{S}) ⊆ Psd(K_{S}) .

Note that the sets T_{S} and cl(T_{S}) depend in general on the choice of S, whereas the set Psd(K_{S}) is uniquely determined by K = K_{S}, independently of the chosen description S.

We fix a group G together with

ϕ : G → GL_{n}(ℝ)

a linear representation. We say that a subset K ⊆ ℝ^{n} is G–invariant if ϕ(g)(K) ⊆ K for every g ∈ G. We can use ϕ to define an action of G on the polynomial ring ℝ[X]:

given p(X) ∈ ℝ[X], define ^{g}p(X) := p(ϕ(g)X) .

Every g ∈ G acts as an ℝ-algebra automorphism of ℝ[X]. In particular, if p(X) ∈ ∑ ℝ[ \underline{X} ]^{2} (i.e. is a sum of squares), then for all g ∈ G, ^{g}p(X) ∈ ∑ ℝ[ \underline{X} ]^{2}. Similarly, if K_{S} is G-invariant and p(X) ∈ Psd(K_{S}), then for all g ∈ G, gp(X) ∈ Psd(K_{S}).

p(X) is said to be G-invariant if for all g ∈ G: ^{g}p(X) = p(X). Note that if K_{S} is defined by invariant polynomials, then K_{S} and Psd(K_{S}) are necessarily G-invariant. For the converse, we cite the following result from [[1]; Cor. 5.4]. Theorem 0.1 Suppose that K = K_{S} is a G-invariant basic closed semi-algebraic set. Then there exists a finite set S′ of G-invariant polynomials such that K_{S′} = K_{S}.

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Preorderings of the ring of invariant polynomials.

Write ℝ[ \underline{X} ]^{G} for the ring of all G-invariant polynomials. We shall always assume that G is a reductive group. So G admits a Reynolds operator. For such groups, Hilbert’s Finiteness Theorem is valid; namely ℝ[ \underline{X} ]^{G} is a finitely generated ℝ-algebra, and the generators may be chosen to be homogeneous polynomials.([12]).

In this talk, for simplicity, we consider the case when G is a finite group. Here, the Reynolds operator is just the average map:

∗ : ℝ[X] → ℝ[ \underline{X} ]^{G}, f ↦ f^{∗} := \frac{1}{|G|} _{g∈G}^{∑} ^{g}f.

From the considerations above, we see that if G finite, then ℝ[X]^{G} is a finitely generated ℝ-algebra. Moreover it has transcendence degree n over ℝ , so is generated by at least n homogeneous invariant polynomials (see [12]).

Note that the Reynolds operator is an ℝ-linear map, which is the identity on ℝ[ \underline{X} ]^{G}, and is a ℝ[ \underline{X} ]^{G}-module homomorphism.

If A ⊆ ℝ[ \underline{X} ] we shall denote by A^{∗} its image in ℝ[ \underline{X} ]^{G} under the Reynolds operator. We note the following important property:

Lemma 0.2 let A ⊆ ℝ[ \underline{X} ]. Assume that A is closed under addition and is (setwise) invariant. then A^{∗} = A ∩ ℝ[ \underline{X} ]^{G}.

For any A ⊆ ℝ[ \underline{X} ], let us denote A^{G} := A ∩ ℝ[ \underline{X} ]^{G}. In particular

(∑ ℝ[ \underline{X} ]^{2})^{G} := (∑ ℝ[ \underline{X} ]^{2}) ∩ ℝ[ \underline{X} ]^{G}

denotes the preordering of ℝ[ \underline{X} ]^{G} of invariant sums of squares.

We now study images of preorderings under the Reynolds operator.

Lemma 0.3 (∑ ℝ[ \underline{X} ]^{2})^{∗} = (∑ ℝ[ \underline{X} ]^{2})^{G}.

Proof: We noted already that if σ ∈ ∑ ℝ[ \underline{X} ]^{2}, then for all g ∈ G,^{g}σ ∈ ∑ ℝ[ \underline{X} ]^{2}. Therefore σ^{∗} ∈ ∑ ℝ[ \underline{X} ]^{2}. This shows that ∑ ℝ[ \underline{X} ]^{2} is setwise invariant. The assertion now follows from Lemma 0.2. □

Let S = {f_{1}, …, f_{k}} ⊂ ℝ[X]^{G}, and K_{S} ⊂ ℝ^{n} the invariant basic closed semialgebraic set defined by S.

We are particularly interested in the following three preorderings of ℝ[X]^{G}, associated to S:

• The preordering of G-invariant psd polynomials

Psd^{G}(K_{S}) := Psd(K_{S}) ∩ ℝ[ \underline{X} ]^{G}

• The preordering

T_{S}^{ℝ[x]^{G}}

in ℝ[X]^{G} which is finitely generated by S.

• The closure cl(T_{S}^{ℝ[ \underline{X} ]^{G}}) of a preordering in ℝ[ \underline{X} ]^{G}:

\{f ∈ ℝ[ \underline{X} ]^{G}; F (f) ≥ 0, F ≠ 0 l. f. on ℝ[ \underline{X} ]^{G} s.t. F (T_{S}^{ℝ[ \underline{X} ]^{G}}) ≥ 0\}.

As before, we have the inclusions

T_{S}^{ℝ[ \underline{X} ]^{G}} ⊆ cl(T_{S}^{ℝ[ \underline{X} ]^{G}}) ⊆ Psd(K_{S})^{G} .

Observe that if T is any preordering in ℝ[ \underline{X} ], then

T^{G} := T ∩ ℝ[ \underline{X} ]^{G}

is a preordering of ℝ[ \underline{X} ]^{G}.

Of course

T_{S}^{ℝ[x]^{G}} ⊆ T_{S}^{G} ⊆ Psd^{G}(K_{S}).

Note that since S is invariant, K_{S}, T_{S} and Psd(K_{S}) are all invariant. Therefore, by Lemma 0.2 we have that

T_{S}^{G} = T_{S}^{∗} and Psd^{G}(K_{S}) = (Psd(K_{S}))^{∗} .

The preordering T_{S}^{G} is easy to describe: Lemma 0.4 T_{S}^{G} is the preordering of ℝ[ \underline{X} ]^{G} generated by (Σℝ[ \underline{X} ]^{2})^{G} and S. Proof: Let h ∈ T_{S}^{G}. Write

h = _{e∈{0,1}^{s}}^{∑} σ_{e}f^{e}, with σ_{e} ∈ ∑ ℝ[X]^{2}

for some {f_{1}, …, f_{k}} ⊆ S. Applying the Reynolds operator we get

h = h^{∗} = (_{e∈{0,1}^{s}}^{∑} σ_{e}f^{e})^{∗} = _{e∈{0,1}^{s}}^{∑} σ_{e}^{∗}f^{e}

(since f_{1}, ⋯, f_{s} ∈ ℝ[ \underline{X} ]^{G}). This is of the required form since σ_{e}^{∗} ∈ (∑ ℝ[ \underline{X} ]^{2})^{G} for each e. □

Remark 0.5 From the Lemma, we see that a set of generators of T_{S}^{G} as a preordering of ℝ[ \underline{X} ]^{G} is of the form S ∪ S_{O} where S_{O} generates the preordering (Σℝ[ \underline{X} ]^{2})^{G} over Σ(ℝ[ \underline{X} ]^{G})^{2} (:=sums of squares of invariant polynomials) . That is

T_{S}^{G} = T_{SUS_{o}}^{ℝ[x]^{G}} with T_{S_{o}}^{ℝ[x]^{G}} = (Σℝ[ \underline{X} ]^{2})^{G} .

In particular,

T_{S}^{G} = T_{S}^{ℝ[x]^{G}} if and only if S_{o} ⊂ T_{S}^{ℝ[x]^{G}} .

In general (Σℝ[ \underline{X} ]^{2})^{G} may properly contain the preordering Σ(ℝ[ \underline{X} ]^{G})^{2} That is

Σ(ℝ[ \underline{X} ]^{G})^{2} ⊆ (Σℝ[ \underline{X} ]^{2})^{G}

but the inclusion may be proper. This was observed in [[2]; Example 5.1], where G-invariant sums of squares have been analyzed.

Example 0.6 Let n = 1 and G = {−1, 1}. We claim that S_{O} = X^{2} generates the preordering (∑ ℝ[ \underline{X} ]^{2})^{G} over Σ(ℝ[ \underline{X} ]^{G})^{2}. Indeed if σ ∈ (∑ ℝ[ \underline{X} ]^{2})^{G}, then

σ = σ^{∗} = _{i}^{∑}(η_{i}^{2})^{∗} with η_{i}(X) ∈ ℝ[X] .

Now (η_{i}^{2})^{∗}(X) = η_{i}^{2}(X) + η_{i}^{2}(−X), so it suffices to prove the claim for η_{i}^{2}(X) + η_{i}^{2}(−X). By separating terms of even and odd degree, we can write

η(X) = μ(X^{2}) + Xθ(X^{2}) ,

with appropriately chosen μ(X), θ(X) ∈ ℝ[X]. Therefore

η_{i}^{2}(X)+η_{i}^{2}(−X) = (μ(X^{2})+Xθ(X^{2}))^{2}+(μ(X^{2})−Xθ(X^{2}))^{2} =

2μ(X^{2})^{2} + 2X^{2}θ(X^{2})^{2}

which is an element of the preordering T_{{X^{2}}}^{ℝ[x]^{G}} as required.

But in general, the preordering (Σℝ[ \underline{X} ]^{2})^{G} need not be finitely generated as we shall now show.

We consider the dyhedral group

G = D_{4} = ⟨a, b|a^{4} = b^{2} = (ab)^{2} = 1⟩

acting on ℝ[x, y] in the standard way:

a (f(x, y)) = f(y, −x), ^{b} (f(x, y)) = f(y, x).

(We note for future reference that ℝ[x, y]^{G} is generated by P_{1}(x, y) := x^{2} + y^{2} and P_{2}(x, y) := x^{2}y^{2}. Note that P_{1} and P_{2} are algebraically independent.)

Example 0.7 The preordering (∑ ℝ[x, y]^{2})^{G} is not finitely generated over ∑(ℝ[x, y]^{G})^{2}.

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Semi-Algebraic Geometry in the Orbit Space.

Let p_{1}, ⋯, p_{k} ∈ ℝ[X] be generators of ℝ[X]^{G}.

Consider the polynomial map

π : ℝ^{n} → ℝ^{k}, \underline{a} = (a_{1}, ⋯, a_{n}) ↦ (p_{1}(\underline{a}) ⋯, p_{k}(\underline{a})) .

By the Tarski-Seidenberg theorem, the image (under this map) of a semi-algebraic set is semi-algebraic. Moreover, since ℝ^{n} is a basic closed semi-algebraic set, so is π(ℝ^{n}) [[1]; Proposition 5.1].

Let ℝ[U]:= the polynomial ring ℝ[U_{1}, ⋯, U_{k}] in k-variables. We fix a finite description v_{1}, ⋯, v_{r} ∈ ℝ[U] of π(ℝ^{n}).

Lemma 0.8 The defining polynomials v_{1}, ⋯, v_{r} ∈ ℝ[U] of π(ℝ^{n}) may be chosen so that π̃^{−1}(v_{1}), ⋯, π̃^{−1}(v_{r}) are all ∈ (∑(ℝ[ \underline{X} ]^{2})^{G}. In this case, for any S ⊂ ℝ[ \underline{X} ]^{G} we have

T_{S}^{G} = T_{S∪{π̃^{−1}(v_{1}),···,π̃^{−1}(v_{r})}}^{G}

For the remaining of the talk, we assume that the finite group G is a generalized reflection group. In this case, ℝ[X]^{G} is generated by k = n algebraically independent elements (see [12]). In the sequel, we shall fix a set of generators p_{1}, ⋯, p_{n}. We let

π̃ : ℝ[X]^{G} → ℝ[U] = ℝ[U_{1}, ⋯, U_{n}]

be the induced ℝ-algebra isomorphism mapping p_{i} to U_{i}. We have

π̃^{−1}(f)(\underline{a}) = f(p_{1}(\underline{a}) ⋯, p_{k}(\underline{a})) for all \underline{a} ∈ ℝ^{n} .

We gather useful properties of the maps π and π̃.

Lemma 0.9 Let S ⊂ ℝ[X]^{G}. We have:

(1) π(K_{S}) = K_{π̃(S)∪{v_{1},···,v_{r}}},

(2) π̃(Psd(K_{S})^{G}) = Psd(π(K_{S})) ⊂ ℝ[U],

(3) π̃(T_{S}^{ℝ[ \underline{X} ]^{G}}) = T_{π̃(S)} ⊂ ℝ[U],

(4) π̃cl(T_{S}^{ℝ[ \underline{X} ]^{G}}) = cl(T_{π̃(S)}) ⊂ ℝ[U]

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Saturation.

Recall that for S ⊂ ℝ[ \underline{X} ]^{G}, T_{S}^{ℝ[ \underline{X} ]^{G}} is saturated means

T_{S}^{ℝ[ \underline{X} ]^{G}} = Psd(K_{S})^{G} .

Similarly T_{S}^{G} is saturated (or T_{S} is G-saturated) means

T_{S}^{G} = Psd(K_{S})^{G} .

Note that T_{S} G-saturated means that every polynomial which is positive semi-definite and invariant is represented in the preordering. Finally, T_{π̃(S)} ⊂ ℝ[U] is saturated means

T_{π̃(S)} = Psd(K_{π̃(S)})

Theorem 0.10 The following are equivalent:

(1) T_{π̃(S)∪π̃(S_{o})∪{v_{1},··· v_{r}}} is saturated,

(2) T_{S∪S_{o}∪\{π̃^{−1}(v_{1}),···,π̃^{−1}(v_{r})\}}^{ℝ[ \underline{X} ]^{G}} is saturated,

(3) T_{S∪{π̃^{−1}(v_{1}),···,π̃^{−1}(v_{r})}}^{G} is saturated.

Remark 0.11 By Lemma 0.8, we may assume that

π̃^{−1}(v_{1}), ⋯, π̃^{−1}(v_{r}) ∈ (∑(ℝ[ \underline{X} ]^{2})^{G} .

In this case, condition (3) of Theorem 0.10 reads: (3’) T_{S}^{G} is saturated.

Applications: We want to apply [4, Theorem 2.2]. We need to define some notions. If K ⊆ ℝ is a non-empty closed semi-algebraic set then K = K_{N}, for N the set of polynomials defined as follows: •If a ∈ K and (−∞, a) ∩ K = ∅, then X − a ∈ N. •If a ∈ K and (a, ∞) ∩ K = ∅, then a − X ∈ N. •If a, b ∈ K, (a, b) ∩ K = ∅, then (X − a)(X − b) ∈ N.

• N has no other elements except these. We call N the natural set of generators for K. We recall [4, Theorem 2.2]:

Theorem 0.12 Assume that K = K_{S} ⊆ ℝ is not compact. Then T_{S} is closed, for any finite set of generators S. Moreover the following are equivalent:

(i) T_{S} is saturated. (ii) T_{S} contains the natural set of generators for K_{S}. (iii) S contains the natural set of generators of K_{S} (up to scalings by positive reals).

Combining Theorem 0.10 with [4, Theorem 2.2] we obtain the following variant of [4, Theorem 2.2]:

Theorem 0.13 Let n = 1 and G as in example 0.6. Let S ⊂ ℝ[ \underline{X} ]^{G}. Assume that K_{S} is non-compact. The following are equivalent:

1. T_{S} is G-saturated,

2. if (a, b), 0 < a < b is a connected component of ℝ \ K_{S} then T_{S} contains (x^{2} −a^{2})(x^{2} −b^{2}), if (−a, a) is a connected component of ℝ\K_{S}, then T_{S} contains x^{2} − a^{2},

3. if (a, b), 0 < a < b is a connected component of ℝ \ K_{S} then S contains (up to a scalar multiple) (x^{2} − a^{2})(x^{2} − b^{2}), if (−a, a) is a connected component of ℝ \ K_{S}, then S contains (up to a scalar multiple) x^{2} − a^{2}.

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G-Invariant Moment Problem.

The general Moment Problem is the following: For a linear functional L on ℝ[X], when is there a positive Borel measure μ or ℝ^{n} such that ∀f ∈ ℝ[X] L(f) = ∫_{ℝ}^{n} fdμ? The following result is due to Haviland.

Theorem 0.14 Given a linear functional L ≠ 0 on ℝ[X] and a closed subset K of ℝ^{n}, one can find a positive Borel measure μ on K such that L(f) = ∫_{K} fdμ (for all f ∈ ℝ[X]) if and only if L(Psd(K)) ≥ 0.

Since Psd(K) is not finitely generated in general([9]), we are interested in approximating it by T_{S}. Therefore, given K a basic closed semi-algebraic subset of ℝ^{n}, we are interested in finding a finite description S ⊂ ℝ[X] of K such that for every L ≠ 0 we have:

L(T_{S}) ≥ 0 implies that L(Psd(K)) ≥ 0 .

If this holds, we say that S solves the K-moment problem. Given K, we say that the K-moment problem is finitely solvable if such an S can be found. So we are searching for a finite description S of K such that every L ≠ 0 which satisfies L(T_{S}) ≥ 0 comes from a positive Borel measure on K = K_{S}. By Theorem 0.14, this is equivalent to finding S such that the following property called (SMP) holds: Psd(K_{S}) = cl(T_{S}).

A linear functional L defined on ℝ[X] is invariant if L is constant on the orbits of the action of G on ℝ[X], that is, if

Ł(f^{∗}) = L(f) for all f ∈ ℝ[X] .

We are interested in the following question:

Let K be an invariant closed subset of ℝ^{n} and Let L ≠ 0 an invariant linear functional defined on the algebra ℝ[ \underline{X} ]. When is there an (invariant) positive Borel measure supported by K such that L(f) = ∫_{K} fdμ for all f ∈ ℝ[ \underline{X} ] ?.

We first note the following:

Lemma 0.15 There is a bijective correspondence between invariant linear functionals on ℝ[ \underline{X} ] and linear functionals on ℝ[ \underline{X} ]^{G}.

Proof: Given L invariant take L | ℝ[ \underline{X} ]^{G}. Conversely, given a linear functional F on ℝ[ \underline{X} ]^{G}, define F^{∗}(f) := F (f^{∗}). Then F^{∗} is invariant linear functionals on ℝ[ \underline{X} ]. Note that the two maps are inverses of each other. □

From this lemma, we see that we can consider linear functionals on ℝ[ \underline{X} ]^{G} instead of invariant linear functionals on ℝ[ \underline{X} ]. We have a G-invariant version of Haviland’s theorem:

Theorem 0.16 Given a linear functional F ≠ 0 on ℝ[X]^{G} and a closed invariant subset K of ℝ^{n}, one can find an invariant positive Borel measure μ on K such that F (f) = ∫_{K} fdμ (for all f ∈ ℝ[X]^{G}) if and only if F (Psd(K)^{G}) ≥ 0.

Proof: If such a measure exists, then clearly F (Psd(K)^{G}) ≥ 0. Now assume that F (Psd(K)^{G} ≥ 0. We claim that F^{∗}(Psd(K)) ≥ 0. Indeed if f ∈ Psd(K) then f^{∗} ∈ Psd(K)^{G} soF^{∗}(f) = F (f^{∗}) ≥ 0. By Theorem 0.14 , F^{∗} is represented by a measure μ supported on K. □

Arguing exactly as before, since Psd^{G}(K) is not finitely generated in general([9]), we are interested in approximating it by T_{S}^{ℝ[ \underline{X} ]^{G}}. Therefore, given K an invariant basic closed semi-algebraic subset of ℝ^{n}, we are interested in finding a finite description S ⊂ ℝ[X]^{G} of K such that for every F ≠ 0 defined on ℝ[ \underline{X} ]^{G} we have:

F (T_{S}^{ℝ[ \underline{X} ]^{G}}) ≥ 0 implies that F (Psd^{G}(K)) ≥ 0 .

If this holds, we say that S solves the invariant Kmoment problem. Given an invariant K, we say that the invariant K-moment problem is finitely solvable if such an S ⊂ ℝ[ \underline{X} ]^{G} can be found.

In other words, given K invariant, we are searching for a finite description S ⊂ ℝ[ \underline{X} ]^{G} of K such that every F ≠ which satisfies F (T_{S}^{ℝ[ \underline{X} ]^{G}}) ≥ 0 comes from an invariant positive Borel measure on K = K_{S}.

By Theorem 0.16, this is equivalent to finding S such that the following property called (ISMP) holds: Psd^{G}(K_{S}) = cl(T_{S}^{ℝ[ \underline{X} ]^{G}}).

Theorem 0.17 Let K ⊂ ℝ^{n} be an invariant basic closed semialgebraic set. Assume that S ⊂ ℝ[ \underline{X} ]^{G} is a finite description of K, i.e. K = K_{S}. Then

S ∪ \{π̃^{−1}(v_{1}), ⋯, π̃^{−1}(v_{r})\} ⊂ ℝ[ \underline{X} ]^{G}

solves the invariant K-moment problem if and only if

π̃(S) ∪ {v_{1}, ⋯, v_{r}} ⊂ ℝ[U]

solves the π(K)-moment problem.

Remark 0.18 (i) In general, we cannot do without {v_{1}, ⋯, v_{r}} in Theorem 0.17. (ii) If however {v_{1}, ⋯, v_{r}} can be chosen so that {v_{1}, ⋯, v_{r}} ⊂ cl (π̃(S)), then indeed S solves the invariant K-moment problem if and only if π̃(S) solves the π(K)-moment problem.

Example 0.19 Let G = {−1, 1} act on ℝ^{n} by

(x_{1}, ⋯, x_{n}) ↦ (−x_{1}, ⋯, x_{n}) .

An argument similar to that given in 0.6 shows that ℝ[ \underline{X} ]^{G} is generated by

p_{1} = X_{1}^{2} , p_{2} = X_{2} ⋯ p_{n} = X_{n} .

Let n = 2, so ℝ[ \underline{X} ]^{G} = ℝ[X^{2}, Y] and ℝ[U] = ℝ[U_{1}, Y], v_{1} = U_{1}. Consider in the XY-plane the invariant subset K_{S} of a cylinder defined by

S = {(X^{2} − 1)(X^{2} − 4) , (1 − Y^{2})} .

Note that π(K_{S}) is again a subset of a cylinder in the U_{1}Y -plane defined by

π̃(S) = {(U_{1} − 1)(U_{1} − 2) , (1 − Y^{2})} .

By [5], π̃(S)∪v_{1} = {U_{1}, (U_{1} −1)(U_{1} −2), (1−Y^{2})} solves the π(K) moment problem. Therefore, by Theorem 0.17, S solves the invariant K-moment problem.

This provides an example of S ⊂ ℝ[ \underline{X} ]^{G} solving the invariant K-moment problem, but not solving the K-moment problem. (Indeed, since the defining inequalities for the boundary of this cylinder are not given by the natural generators, it follows by [5] that S does not solve the Kmoment problem.) In the next Section, we will do better: we provide an example where the K-moment problem is not finitely solvable at all, but the invariant K-moment problem is (see Example 0.22).

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The Averaged Moment Problem.

We now want to analyze the following version of the moment problem concerning representation an invariant linear functionals.

Given K an invariant basic closed semi-algebraic subset of ℝ^{n}, we are interested in finding a finite description S ⊂ ℝ[X]^{G} of K such that for every F ≠ 0 defined on ℝ[ \underline{X} ]^{G} we have:

F (T_{S}^{G})) ≥ 0 implies that F (Psd^{G}(K)) ≥ 0 .

If this holds, we say that S solves the averaged Kmoment problem.

Given an invariant K, we say that the averaged Kmoment problem is finitely solvable if such an S ⊂ ℝ[ \underline{X} ]^{G} can be found. In other words, given K invariant, we are searching for a finite description S ⊂ ℝ[ \underline{X} ]^{G} of K such that every F ≠ 0 which satisfies F (T_{S}^{G}) ≥ 0 comes from an invariant positive Borel measure on K = K_{S}.

By Theorem 0.16, this is equivalent to finding S such that the following property called (ASMP) holds: Psd^{G}(K_{S}) = cl(T_{S}^{G}).

Theorem 0.20 The following are equivalent:

(1) π̃(S)∪π̃(S_{o})∪{v_{1}, ⋯ v_{r}} solves the π(K) moment problem,

(2) S ∪S_{o} ∪{π̃^{−1}(v_{1}), ⋯, π̃^{−1}(v_{r})} solves the invariant K-moment problem

(3) S ∪ {π̃^{−1}(v_{1}), ⋯, π̃^{−1}(v_{r})} solves the averaged Kmoment problem.

Remark 0.21 (i) By Lemma 0.8, we may assume that π̃^{−1}(v_{1}), ⋯, π̃^{−1}(v_{r}) ∈ (∑(ℝ[ \underline{X} ]^{2})^{G}. In this case, condition (3) of Theorem 0.20 reads: (3’) S solves the averaged K-moment problem.

(ii) S_{o} ∪ \{π̃^{−1}(v_{1}), ⋯, π̃^{−1}(v_{r})\} ⊂ cl(T_{S}^{ℝ[ \underline{X} ]^{G}}) then S solves the averaged K-moment problem if and only if S solves the invariant K-moment problem if and only if π̃(S) solves the π(K) moment problem,.

Example 0.22 We reconsider the action of the dyhedral group on the plane ℝ^{2} and on ℝ[x, y]. Let

π : ℝ^{2} → ℝ^{2} (x, y) ↦ (x^{2} + y^{2}, x^{2}y^{2}) , and let
π̃ : ℝ[ \underline{X} ]^{G} → ℝ[u_{1}, u_{2}] p_{1} ↦ u_{1} and p_{2} ↦ u_{2} .

Consider the invariant basic closed semialgebraic set K = K_{S} defined by the inequalities

−1 ≤ (x^{2} − 1)(y^{2} − 1) ≤ 0

i.e. by S ⊂ ℝ[ \underline{X} ]^{G}:

S = {(x^{2} − 1)(y^{2} − 1) + 1, −(x^{2} − 1)(y^{2} − 1)} .

Then π(ℝ^{2}) is defined by

v_{1} := u_{1} ≥ 0 , v_{2} := u_{2} ≥ 0 , v_{3} := u_{1}^{2} − 4u_{2} ≥ 0 .

Computing π(K_{S}) ⊂ ℝ^{2} we find

π(K_{S}) = K_{π̃(S)∪{v_{1},v_{2},v_{3}}} .

Now

π̃((x^{2} − 1)(y^{2} − 1)) = u_{2} − u_{1} + 1

so π(K_{S}) is defined by the inequalities

u_{1} − 2 ≤ u_{2} ≤ u_{1} − 1 .

We see that π(K_{S}) is a cylinder with compact cross-section (actually, a convex polyhedron defined by linear inequalities). So

π̃(S) ∪ {v_{1}, v_{2}, v_{3}}

solves the π(K)-moment problem, by [4]. By Theorem 0.17, we get that

S ∪ {π̃^{−1}(v_{1}), π̃^{−1}(v_{2}), π̃^{−1}(v_{3})}

solves the invariant K_{S}-moment problem. Computing, we get π̃^{−1}(v_{1}) = x^{2}+y^{2} , π̃^{−1}(v_{2}) = x^{2}y^{2} , and π̃^{−1}(v_{3}) = (x^{2}−y^{2})^{2}, which are all invariant sums of squares. We conclude that S solves the averaged K_{S}- moment problem . We now claim that the moment problem for K is not finitely solvable. This can be established by applying [[6];Corollary 3.10].

The End References

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